@LeakyNun Well I know a bit about ultrafilters, namely the bit that is sufficient for constructing ultrapowers.

Or more accurately I knew and can re-know if needed. =P

can you develop Tychonoff to me?

@LeakyNun I did not learn about Tychonoff's theorem. Did you already understand how the proof works assuming you have ultrafilters?

I don't understand ultrafilters, so no

Hmm, the part I can explain is ultrafilters. Basically you just keep extending (via transfinite induction) any given filter on S until it is maximal. Then you end up with a filter that includes either X or S\X for each X⊆S.

Alternatively, you use Zorn's lemma on the partial order of filters on S.

A filter is a collection of subsets of S that is closed under intersection and superset.

You can think of a filter as a specific notion of largeness, where two large subsets have a large intersection, and anything larger than a large subset is also large.

what is the filter definition of compact and why is it true?

if we have a model (M,r) of ZFC where r is well-founded, then we have a transitive model of ZFC right

@LeakyNun I don't know how filters are related to Tychonoff's theorem, so I probably can't answer that without first learning enough about the latter.

alright

@LeakyNun Yes we can use Mostowski collapse to obtain a transitive model. The original may not be transitive.

let's say I know about ultraproducts

then how do you prove compactness theorem?

Let F be the set of finite subsets of the theory S. Let X[T] be the set of all elements of F that contain (⊇) T. Let D be { Y : Y ⊆ F and Y ⊇ X[T] for some T in F }. Then D is a filter on F. (Closure under superset is trivial. Closure under intersection holds since X[T1]⋂X[T2] ⊆ X[T1⋂T2].) Let U be an ultrafilter on F extending D. Let M[T] be a model of T for each T in F (using AC). Let M* be the ultraproduct of M[T] over U. Then M* satisfies every T in F since X[T] is in U.

But it is much simpler if S is countable. You just enumerate S as Q[1..], and let M[k] be a model of Q[1..k], and let U be an ultrafilter on N extending the Frechet filter (all cofinite subsets of N), and let M* be the ultraproduct of M[k] over U. Then every Q[k] is satisfied by M[k..] and hence also by M*.

But in my opinion it's silly to use ultraproducts when all you actually need is a well-ordering of S. The ultraproduct needs a well-ordering of P(F), which is much more 'unreal' especially when S is countable.

@LeakyNun: Got to go. See you!

@user21820

When we say "given any function f from R to R we say that f is monotone iff forall x,x' in R if x < x' then f(x) < f(x')" we abuse notation, right? on one side we are talking about the function as if it has been constructed from sets and on the other side we use f(x) as if it was developed from a definatorial expansion.

It should've been more like "given any function F from R to R we say that F is monotone iff forall x,x' in R if x < x' then f(x) < f(x')"